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Four charges are at the corners of a square of side a as show in the figure. a) Determine the magnitude and direction of the electric field at the location of charge q. b) What is the resultant force on q?

[/PLAIN] [Broken]

I made E1 equal the electric field of 2q and q. I made E2 equal the electric field of 3q and q. I made E3 equal the electric field of 4q and q. So, when I wrote out the vector notations of each electric field, I got:

[tex] E_{1} = k_{e} \frac {2q} {a^2} \hat {i} [/tex]

[tex] E_{3} = k_{e} \frac {4q} {a^2} \hat {j} [/tex]

[tex] E_{2} = k_{e} \frac {3q} {a^2} [/tex]

How would u break down E2 into x and y components?

Also, after getting E1, E2, and E3, E = E1 + E2 + E3. So, you add all the vectors up and a x and y component. Next, magnitude of E = square root of x component squared plus y component squared. The angle is tan of y component over x component.

Did I do it right for part A? Any help would be great! thx !!

Code:

`[PLAIN]http://east.ilrn.com/graphing/bca/user/appletImage?dbid=1131113015`

I made E1 equal the electric field of 2q and q. I made E2 equal the electric field of 3q and q. I made E3 equal the electric field of 4q and q. So, when I wrote out the vector notations of each electric field, I got:

[tex] E_{1} = k_{e} \frac {2q} {a^2} \hat {i} [/tex]

[tex] E_{3} = k_{e} \frac {4q} {a^2} \hat {j} [/tex]

[tex] E_{2} = k_{e} \frac {3q} {a^2} [/tex]

How would u break down E2 into x and y components?

Also, after getting E1, E2, and E3, E = E1 + E2 + E3. So, you add all the vectors up and a x and y component. Next, magnitude of E = square root of x component squared plus y component squared. The angle is tan of y component over x component.

Did I do it right for part A? Any help would be great! thx !!

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